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Both sides previous revision Previous revision Next revision | Previous revision | ||
184_notes:examples:week8_wheatstone [2021/06/29 00:15] – schram45 | 184_notes:examples:week8_wheatstone [2021/07/22 18:28] (current) – schram45 | ||
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[{{ 184_notes: | [{{ 184_notes: | ||
- | To find $\Delta V_{\text{light}}$, | + | To find $\Delta V_{\text{light}}$, |
[{{ 184_notes: | [{{ 184_notes: | ||
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Applying the Node Rule and the Loop Rule, we obtain the following equations: | Applying the Node Rule and the Loop Rule, we obtain the following equations: | ||
\begin{align*} | \begin{align*} | ||
- | I &= I_1 + I_2 & | + | I &= I_1 + I_2 & |
- | I &= I_3 + I_4 & | + | I &= I_3 + I_4 & |
\Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 & | \Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 & | ||
\Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 & | \Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 & | ||
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Since we know $\Delta V_3$ and all the resistances, | Since we know $\Delta V_3$ and all the resistances, | ||
$$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$ | $$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$ | ||
+ | |||
+ | Looking at loop 2 alone between the power source and resistor 3 we would expect the voltage across any other elements in that loop to be small. Our answer agrees with this observation as 2.37 is quite small compared to both the battery and resistor 3. | ||